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Chapter 10: Problem 33

In a 2015 study reported in the Journal of American College Health, Cho et al. surveyed college students on their use of apps to monitor their exercise and fitness. The data are reported in the table. Test the hypothesis that fitness app use and gender are associated. Use a \(0.05\) significance level. See page 552 for guidance. $$ \begin{array}{lcc} \text { Use } & \text { Male } & \text { Female } \\ \text { Yes } & 84 & 268 \\ \text { No } & 9 & 57 \end{array} $$

Short Answer

Expert verified
The solution will depend on the computed Chi-square statistic and its comparison to the critical Chi-Square value.

Step by step solution

01

Set up the Hypotheses

Based on the exercise, the null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)) are as follows:\n\nNull hypothesis (\(H_0\)): Fitness app use and gender are not associated, they are independent.\n\nAlternative hypothesis (\(H_a\)): Fitness app use and gender are associated, they are not independent.
02

Calculate the Expected Frequencies

The expected frequency for each cell in the contingency table is calculated as follows:\n\n(Expected frequency) = (Row Total * Column Total) / Grand Total\n\nHere are the calculated expected frequencies for each cell:\n\nMale who uses app: \( (84+9)*(84+268) / (84+268+9+57) = 135.36 \)\n\nFemale who uses app: \( (268+57)*(84+268) / (84+268+9+57) = 216.64 \)\n\nMale who doesn't use app: \( (84+9)*(9+57) / (84+268+9+57) = 57.64 \)\n\nFemale who doesn't use app: \( (268+57)*(9+57) / (84+268+9+57) = 8.36 \)
03

Calculate Chi-Square Test Statistic

The formula for calculating the Chi-Square test statistic is:\n\n \(\chi^2 = \Sigma [ (O-E)^2 / E ] \)\n\nWhere 'O' represents the observed frequency and 'E' represents the expected frequency. After calculating, we will have a single \(\chi^2\) value.
04

Find Critical Value and Make Decision

To make a decision about the significance of the results, we compare our calculated \(\chi^2\) value to the critical value from the \(\chi^2\) distribution with degree of freedom \((C-1) * (R-1)\), where 'C' is the number of columns and 'R' is the number of rows. If our calculated \(\chi^2\) value is greater than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
05

Interpret the Results

Our decision should be stated in terms of the problem. If we fail to reject the null hypothesis, we can say that there is no strong evidence that fitness app use and gender are associated. If we reject the null hypothesis, we can say that the data suggest there is an association between fitness app use and gender.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
When you hear about a null hypothesis, think of it as the starting point in a statistical test. It is a default statement that there is no effect or no difference; in the context of our exercise, it's the claim that fitness app usage is unrelated to gender among college students. Formally, we've expressed it as 'Fitness app use and gender are not associated'.

Researchers use the null hypothesis as a basis for comparison against experimental data. When conducting a Chi-Square Test, as in our exercise, rejecting the null hypothesis indicates that there is enough evidence in the data to support a relationship between the variables studied鈥攊n this case, gender differences in the use of fitness apps. If the null hypothesis is not rejected, it implies that any observed differences in the data might be due to random chance rather than a real effect.
Expected Frequency
The expected frequency is a crucial element in a Chi-Square Test, providing a benchmark to compare against the observed data. It's a projection of how many instances should fall into each category of the table if the null hypothesis were true鈥攖hat is, if there's no association between the variables.

To get these numbers, you apply a specific formula: (Row Total * Column Total) / Grand Total. In our exercise, we made these calculations for males and females, both users and non-users of fitness apps. Expected frequencies are not based on the actual data collected but on the distribution we'd expect to see by chance alone. They're not actual counts but values derived from the overall sample's structure.
Statistical Significance
The term statistical significance acts as a judge in the courtroom of scientific research. It's a measure of whether your results are likely due to something other than random chance. The cut-off point for this judgment is known as the significance level, commonly set at 0.05 or 5%. It's like setting a strict rule that only when something has less than a 5% probability of being a fluke will you take it seriously.

In our study, assessing whether fitness app usage is gender-dependent requires us to calculate a Chi-Square statistic and compare it to a critical value associated with our chosen significance level. If our calculated statistic is high enough to cross the threshold defined by the significance level, it's like saying, 'This didn't happen by accident; something's going on here,' and so we conclude that the results are statistically significant. This understanding helps to reveal the actual import of the data, moving beyond the numbers to grasp the study's implications.

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Most popular questions from this chapter

In a 2018 article published in The Lancet, Sprigg et al. studied the effect of tranexamic acid in treating patients with intracerebral hemorrhages using a randomized, placebo-controlled trial. Of the 1161 subjects treated with tranexamic acid, 383 suffered an adverse outcome after 2 days. Of the 1164 subjects given a placebo, 419 suffered an adverse outcome after 2 days. a. Find the percentage in each group that suffered an adverse outcome. Round off to one decimal place as needed. b. Create a two-way table with the treatment labels (drug/placebo) across the top. c. Test the hypothesis that treatment and adverse outcome are associated using a significance level of \(0.05\).

In 2017 the Pew Research Center published a report on the demographics of the U.S. military. The following table shows the ethnic breakdown of active-duty U.S. military services and the ethnic breakdown of the U.S. population. Would it be appropriate to use the data in the table to conduct a chisquare goodness of fit test to determine if the ethnic breakdown of the military matches the ethnic distribution of the U.S. population? If so, do the test. If not, explain why it would be inappropriate to do so. $$ \begin{array}{lcc} \text { Ethnicity } & \text { \% Active-Duty Military } & \text { \% U.S. Population } \\ \hline \text { White } & 60 & 61 \\ \text { Black } & 17 & 12 \\ \hline \text { Hispanic } & 12 & 18 \\ \hline \text { Asian } & 4 & 6 \\ \hline \text { Other } & 7 & 3 \\ \hline \end{array} $$

Suppose you randomly assign some parolees to check in once a week with their parole officers and others to check in once a month, and observe whether they are arrested within 6 months of starting parole.

In Chapter 9 , you learned some tests of means. Are tests of means used for numerical or categorical data?

The following table shows the average number of vehicles sold in the United States monthly (in millions) for the years 2001 through 2018 . Data on all monthly vehicle sales for these years were obtained and the average number per month was calculated. Would it be appropriate to do a chi-square analysis of this data set to see if vehicle sales are distributed equally among the months of the year? If so, do the analysis. If not, explain why it would be inappropriate to do so. (Source: www.fred.stlouisfed.org) $$ \begin{array}{|l|l|} \hline \text { Month } & \text { Avg Sales per Month (in millions) } \\ \hline \text { Jan } & 15.7 \\ \hline \text { Feb } & 15.7 \\ \hline \text { Mar } & 15.8 \\ \hline \text { Apr } & 15.8 \\ \hline \text { May } & 15.8 \\ \hline \text { June } & 15.7 \\ \hline \text { July } & 16.1 \\ \hline \text { Aug } & 16.1 \\ \hline \text { Sept } & 15.8 \\ \hline \text { Oct } & 15.9 \\ \hline \text { Nov } & 15.9 \\ \hline \text { Dec } & 15.9 \\ \hline \end{array} $$
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